Simplifying powers of
i

The powers of
i are cyclic. Let’s look at what happens when we raise
i to increasing powers.

i1=ii2=−1i3=i2⋅i=−1⋅i=−ii4=i3⋅i=−i⋅i=−i2=−(−1)=1i5=i4⋅i=1⋅i=i

We can see that when we get to the fifth power of
i, it is equal to the first power. As we continue to multiply
i by increasing powers, we will see a cycle of four. Let’s examine the next four powers of
i.

i6=i5⋅i=i⋅i=i2=−1i7=i6⋅i=i2⋅i=i3=−ii8=i7⋅i=i3⋅i=i4=1i9=i8⋅i=i4⋅i=i5=i

The cycle is repeated continuously:
i,−1,−i,1, every four powers.

Simplifying powers of
i

Evaluate:
i35.

Since
i4=1, we can simplify the problem by factoring out as many factors of
i4 as possible. To do so, first determine how many times 4 goes into 35:
35=4⋅8+3.

As we saw in
[link] , we reduced
i35 to
i3 by dividing the exponent by 4 and using the remainder to find the simplified form. But perhaps another factorization of
i35 may be more useful.
[link] shows some other possible factorizations.

Factorization ofi35

i34⋅i

i33⋅i2

i31⋅i4

i19⋅i16

Reduced form

(i2)17⋅i

i33⋅(−1)

i31⋅1

i19⋅(i4)4

Simplified form

(−1)17⋅i

−i33

i31

i19

Each of these will eventually result in the answer we obtained above but may require several more steps than our earlier method.

Media

Access these online resources for additional instruction and practice with complex numbers.